Interaction of the three kinds

Question

There're three kinds of it.

• Any two of the same kind will give you that same kind.

• Any two of the different kinds will give you the remaining kind.

What is it?

I don't know the answer. I'm trying to find a real-world embodiment for this concept.

---

Since there was a lot of misunderstanding I'll try to elaborate. Formally, we have three types (classes, families, kinds) of something, denote them $A,B$ and $C$. We can take instances of those types, like $a$ of $A$ ($a\in A$ as a notation), where $a$ is some specific object that is assosiated with the class $A$. We also have a mapping (combining, mixing, colliding) procedure $\times$ that for any $a_1,a_2\in A$, $b_1,b_2\in B$, $c_1,c_2\in C$:

$\qquad a_1\times a_2\in A,\quad b_1\times b_2\in B,\quad c_1\times c_2\in C$

$\qquad a_1\times b_1\in C,\quad b_1\times c_1\in A,\quad c_1\times a_1\in B$

The goal is to find a natural embodiment of these types, instances and mapping procedure that behaves by these rules without us having to explicitly explain them.

We can make up a bunch of forced examples, like the mechanics of some objects in a video game, or a part of some logic mechanism or schematics. But to construct such an example, we would actually need to lay out the rules once again. They don't emerge naturally and that's the problem.

Answer 1

As pointed out by FrodCube in the comment, the structure cannot be a group. The structure also cannot be a ring, since by your requirement, $(x*x)*y = x*y = z \neq y = x*z = x*(x*y)$, so the operation is not associative, which is required by ring.

A mathematical structure that satisfies the two conditions is the structure with three elements $\{0, 1, 2\}$ and the operation: $*$ such that:

$x*y = -(x+y) \mod 3$

Since:

Condition 1:
$0 * 0 = -(0 + 0) \mod 3 = 0$
$1 * 1 = -(1 + 1) \mod 3 = -2 \mod 3 = 1$
$2 * 2 = -(2 + 2) \mod 3 = -4 \mod 3 = 2$

Condition 2:
$0 * 1 = -(0 + 1) \mod 3 = -1 \mod 3 = 2$
$1 * 2 = -(1 + 2) \mod 3 = -3 \mod 3 = 0$
$2 * 0 = -(2 + 0) \mod 3 = -2 \mod 3 = 1$
Also they are commutative, so $x*y=y*x$

Note that the "elements" above can serve as categories also, for example:

Let
$A = \{\ldots, -6, -3, 0, 3, 6,\ldots\},$
$ B=\{\ldots, -5, -2, 1, 4, 7, \ldots\},$
$C=\{\ldots, -4, -1, 2, 5, 8, \ldots\}$.
The operation $\times$ defined as $x\times y = -(x+y)$ will satisfy the formal requirements specified in the question.

If this is "real enough" for you as embodiment, then that's great. Otherwise we need to find "more real" example, maybe something like this:

- Let's say we have three types of cards X, Y, Z. Then we gain a combo when either we have three of the same kind or one from each kind. Then the operation you are looking for would be "the third card such that we have a combo given the first two cards". Like f(X, X) = X (to get the "three of the same kind" combo), f(X, Y) = Z (to get the "one from each kind" combo), etc.
Some commenters point out that the game Risk or Set might be some contexts where this operation can take place in real life.

For some other ideas:

- Some form of Scissor-Paper-Stone may work?
- Take some ideas from here.

Answer 2

Draw three lines intersecting at 60 degree angles:

The reflection of any line over itself is itself. The reflection of any line over any other line is the third line.

Answer 3

Partial answer

I guess there is
a + b = c
a + c = b
b + c = a

Plus:
a + a = a
b + b = b
c + c = c

Which is only true for a = b = c = 0 leading to only one kind. Correct me if I'm wrong.

---

Real-world embodiment:
1. no time
2. a lot of money
3. a lot of work

The combinations are working from my point of view but I'm not sure if this is subjective.

Answer 4

This partly fits, in that you can derive C from just A and B...

Take three vectors with sum zero, A, B and C.

Taking two vectors from the three (that can be the same or not the same), and summing those two vectors, you would get:

2A from A and A
2B from B and B
2C from C and C
-A from B and C
-B from A and C
-C from A and B

So:

• Any two of the same kind will give you TWICE that same kind
• Any two of the different kinds will give you the OPPOSITE OF the remaining kind

Edit: Of course, this isn't that special for vectors, it works for any numbers! If three numbers add up to 0, you can find the third from the first two...

Answer 5

Is it flag with colors 'blue, yellow and green'?

same kind will give you that same kind
color mixed with itself gives same color
Two of the different kinds will give you the third kind
Blue + Yellow = Green

Answer 6

Partial fit -

It can be rectangular coordinates (x,y,z) and the operation be vector product.

two of the same kind will give you that same kind

x X x=y Xy=z X z= 0.

two of the different kinds will give you the remaining kind

x X y=z ,y X z=x ,zXx =y

Answer 7

Not a real world example, but taking
Group A: vectors of form $(x,0,0)$
Group B: vectors of form $(0,x,0)$
Group C: vectors of form $(0,0,x)$

and Vector multiplication (the cross product) as the "interaction".

Edit: Cross product gives a vector perpendicular to both the operands, so $a \times b \in C$
but after rechecking, $a \times a = 0$... which makes this answer wrong.

Answer 8

I won't hide my answer because you are actually looking for a solution. Others may see this and come up with something better.

My answer is based upon the Ideal Gas Law:

$$PV=nRT$$

Where $P = Pressure$, $V = Volume$, and $T=Temperature$. Assume the others are constants.

Let $A=\Delta{T}$, then $\Delta{T}$ plus another $\Delta{T}$ produces an overall $\Delta{T}$. $A+A=A$

Let $B=\Delta{V}$, then $\Delta{V}$ plus another $\Delta{V}$ produces an overall $\Delta{V}$. $B+B=B$

Let $C=\Delta{P}$, then $\Delta{P}$ plus another $\Delta{P}$ produces an overall $\Delta{P}$. $C+C=C$

This produces the following based on the Ideal Gas Law equation:

$\Delta{T}$ + $\Delta{V}$ produces $\Delta{P}$ based on $P=\frac{nRT}{V}$. $A + B=C$

$\Delta{T}$ + $\Delta{P}$ produces $\Delta{V}$ based on $V=\frac{nRT}{P}$. $A + C=B$

$\Delta{V}$ + $\Delta{P}$ produces $\Delta{T}$ based on $T=\frac{PV}{nR}$. $B + C=A$

In other words, "It" would be "Change". Like changes result in a like overall change. Different changes result in yet another change.

This is the best I can imagine right now. Maybe someone can find another triple-variable equation.

Answer 9

Two points in a finite projective plane determine a line. If the finite projective plane is taken to be over the field of three elements, then each coordinate of the operation that takes two points and produces the third point that is colliear with the two points has the desired property.